Retained inertia & jump

[DFW]

After looking over this question. For a House rule I decided that a ship does NOT retain inertia in Jump Space. A ship leaving our universe loses its inertial reference (see Mach's principle) and loses its inertia related to our universe.

It does, while in J-space, get a new inertia frame. So, that when it is dumped back, it is moving in conjunction with the target systems frame of reference motion. EI: that of the system star.

 

[barnest2]

So it is at zero in relation to the star?

 

[DFW]

So it is at zero in relation to the star?

Yes. So if you are aiming ahead of the target planet, it is coming up on you at its orbital speed. Of course, you have to start thrusting as you will begin to fall towards the star. So you aim a little ahead of the orbital path and little further out.

 

[barnest2]

Sounds good to me. It's what I've been using.

 

[DFW]

Sounds good to me. It's what I've been using.

Cool. It also does away with the killer asteroid being jumped into a system and vaporizing a planet.

 

[barnest2]

Sounds good to me. It's what I've been using.

Cool. It also does away with the killer asteroid being jumped into a system and vaporizing a planet.

Indeed. But of course you can still set up outside the system and then accelerate for 15 days or so.

 

[DFW]

Sounds good to me. It's what I've been using.

Cool. It also does away with the killer asteroid being jumped into a system and vaporizing a planet.

Indeed. But of course you can still set up outside the system and then accelerate for 15 days or so.

Yes but, if it is a planet of average interstellar TL, they will see it a LONG way off. It would be intercepted and diverted using large nukes.

 

[barnest2]

Yay Nukes :roll:

 

[rust]

Yes but, if it is a planet of average interstellar TL, they will see it a LONG way off. It would be intercepted and diverted using large nukes.

No need to waste a nuke. If it is detected that early, the maneuver drive
of a launch, running for a week or so, would be sufficient to change the
asteroid's vector enough to turn it into a not so near miss.

 

[DFW]

Yay Nukes :roll:

ALWAYS good to have around. :lol:

 

[far-trader]

Yes but, if it is a planet of average interstellar TL, they will see it a LONG way off. It would be intercepted and diverted using large nukes.

No need to waste a nuke. If it is detected that early, the maneuver drive
of a launch, running for a week or so, would be sufficient to change the
asteroid's vector enough to turn it into a not so near miss.

IF someone is going to the trouble of throwing near-c/fractional-c rocks (yes plural, there won't be just one) at you they'll have defenses set up to deal with your attempts to stop them. Sending a single launch won't be enough. Sending a fleet of warships might be enough for your rock killer to get through. If you have a fleet of warships in the system at the time.

Better to simply apply a little reality and common sense and treat maneuver drives in a more limited fashion, as they were intended to be.

 

[DFW]

Better to simply apply a little reality and common sense and treat maneuver drives in a more limited fashion, as they were intended to be.

Well, per every movement rule in Trav, (see travel time tables) they are intended to let you continuously accelerate at LEAST from Earth to distant GG's at 6G's.

So, even at those canonical rates, a "rock" gets moving fast enough to pretty much devastate a planet.

 

[far-trader]

Better to simply apply a little reality and common sense and treat maneuver drives in a more limited fashion, as they were intended to be.

Well, per every movement rule in Trav, (see travel time tables) they are intended to let you continuously accelerate at LEAST from Earth to distant GG's at 6G's.

Half actually, you forgot the turn-over and decelerate act.

True, it's still fast and will do damage, but as you note you'll be able to spot it and the valued systems will have resources to deal with it.

I'm just saying you don't need to drop the retained vector to prevent near-c rocks popping out of jump space. I'm not saying your idea isn't a nice one, but we do have canon on retained velocity as well. While we have no canon on near-c or even high velocity rocks used as WMDs which says something.

 

[rust]

I'm just saying you don't need to drop the retained vector to prevent near-c rocks popping out of jump space.

This would be nonsensical anyway. Since the arrival cannot be timed pre-
cisely the chance that the rock would emerge from jump space on a vec-
tor taking it to the target planet would be almost zero, and if it is at near
c already, it would take an incredible amount of energy to alter its vector.

 

[GypsyComet]

Well, per every movement rule in Trav, (see travel time tables) they are intended to let you continuously accelerate at LEAST from Earth to distant GG's at 6G's.

The usual is 10,000 stellar diameters, IIRC. In our system that's all four gas giants, Pluto, Haumea, Makemake, and Eris. A looong way out.

if it's 1,000 stellar diameters, that'll only reach Saturn.

 

[GypsyComet]

Since the arrival cannot be timed pre-
cisely the chance that the rock would emerge from jump space on a vec-
tor taking it to the target planet would be almost zero.

Which is why you use ten or a hundred smaller ones instead, or jump one with a fracturing charge that goes off after emergence, shotgunning whatever region of the target system it moves through. Do this multiple times, only stopping when they send a converted wet navy battleship after you.

 

[far-trader]

I'm just saying you don't need to drop the retained vector to prevent near-c rocks popping out of jump space.

This would be nonsensical anyway. Since the arrival cannot be timed pre-
cisely the chance that the rock would emerge from jump space on a vec-
tor taking it to the target planet would be almost zero, and if it is at near
c already, it would take an incredible amount of energy to alter its vector.

IMO the randomness is the nonsensical part That and the idea that wasting several millions of credits putting a jump drive on a rock as a weapon was ever a brilliant strategy.

 

[rust]

That and the idea that wasting several millions of credits putting a jump drive on a rock as a weapon was ever a brilliant strategy.

It is even worse than that. Just some quick calculations – I hope with-
out errors.

A rock big enough to survive the impact with a planet's atmosphere
should probably have a diameter of at least 100 meter, which would
give it a volume of 520,000 cubic meters. Assuming it has the densi-
ty of granite and therefore a mass of 2.7 tons per cubic meter it would
have a total mass of approximately 1.4 million tons.

Using the GURPS Traveller formula that a Traveller starship has an
average mass of 5 tons per dton, this would make the rock the equi-
valent of a starship of 280,000 dtons. To accelerate it with 1 G would
require a maneuver drive of 2,800 dtons at a cost of 1,400 MCr.

1,400 MCr per rock at a hit chance of almost zero would not exactly
be a prudent investment - and it does not even include the jump drive
required to transport the rock into another system ...

 

[far-trader]

That and the idea that wasting several millions of credits putting a jump drive on a rock as a weapon was ever a brilliant strategy.

It is even worse than that...

Did I type millions? I was going to type billions but wasn't sure. Thanks for the quick check rust

Yeah, pretty silly budgetary wise. Though you could keep the maneuver drive and power plant for reuse by making it a separate ship that just pushed the rock up to speed. You still sacrifice the jump drive though...

 

[rust]

You still sacrifice the jump drive though...

For 520,000 cubic meters = 38,500 dtons volume you can get the jump
drive for only another ca. 1,500 MCr. 8)